Question

QUESTION 8

The next three questions are based on a synthesis performed by Doc Inmaking. He is doing a common organic chemistry test for C=C groups. He starts with 0.6734 g of hexene, C₆H₁₂, and reacts it with excess Br₂. The Br₂ adds to the C=C group to form the product C₆H₁₂Br₂ according to this chemical equation.

C₆H₁₂(l) + Br₂(l) → C₆H₁₂Br₂(l)

Predict Doc's theoretical yield of C₆H₁₂Br₂ product. (Answer with four significant digits, units of g)

1 points   

QUESTION 9

Doc collects 1.357 g of his C₆H₁₂Br₂ product. What is his percent yield? (Answer with four significant digits, and units of %)

1 points   

QUESTION 10

Based on the answers to the previous two questions, what is Doc's percent error for his synthesis experiment? (Answer to four significant digits, and units of %)

1 points   

QUESTION 11

The next three questions are based on the combustion of propane gas (C₃H₈) with oxygen to produce carbon dioxide and water vapor.

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)

Stu Dent measures exactly 5.10 g of C₃H₈ and burns it with excess oxygen gas. What is his theoretical yield of carbon dioxide gas?  What is his theoretical yield of water vapor?  (Both answers should have 3 significant digits and units of g)

2 points   

QUESTION 12

Stu measures his experimental products. He finds 13.3 g CO₂ and 6.94 g H₂O formed. What is the percent yield for carbon dioxide?  What is the percent yield for water?  (Both answers should have 3 significant digits with units of %)

2 points   

QUESTION 13

Calculate Stu's percent error for his two products. (Both answers should have 3 significant digits with units of %)  

Percent error for carbon dioxide?  Percent error for water?

Answer 1

C₆H₁₂(l) + Br₂(l) → C₆H₁₂Br₂(l)

1 mole of hexane react with Br2 to gives 1 mole of C6H12Br2

84g of hexane react with Br2 to gives 244g of C6H12Br2

0.6734g of hexane react with br2 to gives = 244*0.6734/84   = 1.96g

Theoritical yield of C6H12Br2 = 1.96g

percentage yiled = 1.357*100/1.96   = 69.34%

percentage error   = 1.96-1.357*100/1.96   = 30.76%

11.

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)

1 mole of C3H8 combustion to gives 4 moles of H2O

44g of C3H8 combustion to gives 4*18g of H2O

5.1g of C3H8 combustion to gives = 4*18*5.1/44   = 8.35g of H2O

percentage yiled of H2O   = 6.94*100/8.35   = 83.11%

1 mole of C3H8 combustion to gives 3 moles of CO2

44g of C3H8 combustion to gives 3*44g of CO2

5.1g of C3H8 combustion to gives = 3*44*5.1/44   = 15.3g of CO2

percentage yield of CO2 = 13.3*100/15.3   = 86.93%